Question: Multiply the following complex numbers: $({3+3i}) \cdot ({4+i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3+3i}) \cdot ({4+i}) = $ $ ({3} \cdot {4}) + ({3} \cdot {1}i) + ({3}i \cdot {4}) + ({3}i \cdot {1}i) $ Then simplify the terms: $ (12) + (3i) + (12i) + (3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 12 + (3 + 12)i + 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 12 + (3 + 12)i - 3 $ The result is simplified: $ (12 - 3) + (15i) = 9+15i $